Answer:
3. [tex]\dfrac{4}{3 \pi}[/tex]
Step-by-step explanation:
Volume of a cone:
[tex]V=\dfrac13\pi r^2h[/tex]
(where r is the radius and h is the height)
Given:
[tex]\implies r=\dfrac14h[/tex]
Therefore, substitute [tex]r=\frac14h[/tex] into the volume formula to find the volume of the cone in terms of h:
[tex]\implies V=\dfrac13\pi \left(\dfrac14h\right)^2h[/tex]
[tex]\implies V=\dfrac{1}{48}\pi h^3[/tex]
Differentiate with respect to h:
[tex]\implies \dfrac{dV}{dh}=3 \cdot \dfrac{1}{48} \pi h^2=\dfrac{\pi h^2}{16}[/tex]
Volume is increasing at a constant rate of 3cm/s:
[tex]\implies \dfrac{dV}{dt}=3[/tex]
Use the chain rule to find [tex]\dfrac{dh}{dt}[/tex]
[tex]\implies \dfrac{dh}{dt}=\dfrac{dV}{dt}\times\dfrac{dh}{dV}[/tex]
[tex]\implies \dfrac{dh}{dt}=3\times\dfrac{16}{\pi h^2}[/tex]
[tex]\implies \dfrac{dh}{dt}=\dfrac{48}{\pi h^2}[/tex]
When h = 6:
[tex]\implies \dfrac{dh}{dt}=\dfrac{48}{\pi \cdot 6^2}=\dfrac{4}{3 \pi}[/tex]
